3.174 \(\int \frac{\cos ^3(c+d x) (A+A \sec (c+d x))}{(a-a \sec (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=236 \[ \frac{85 A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a-a \sec (c+d x)}}\right )}{8 a^{3/2} d}-\frac{15 A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a-a \sec (c+d x)}}\right )}{\sqrt{2} a^{3/2} d}+\frac{35 A \sin (c+d x)}{8 a d \sqrt{a-a \sec (c+d x)}}+\frac{4 A \sin (c+d x) \cos ^2(c+d x)}{3 a d \sqrt{a-a \sec (c+d x)}}-\frac{A \sin (c+d x) \cos ^2(c+d x)}{d (a-a \sec (c+d x))^{3/2}}+\frac{25 A \sin (c+d x) \cos (c+d x)}{12 a d \sqrt{a-a \sec (c+d x)}} \]

[Out]

(85*A*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a - a*Sec[c + d*x]]])/(8*a^(3/2)*d) - (15*A*ArcTan[(Sqrt[a]*Tan[c + d
*x])/(Sqrt[2]*Sqrt[a - a*Sec[c + d*x]])])/(Sqrt[2]*a^(3/2)*d) - (A*Cos[c + d*x]^2*Sin[c + d*x])/(d*(a - a*Sec[
c + d*x])^(3/2)) + (35*A*Sin[c + d*x])/(8*a*d*Sqrt[a - a*Sec[c + d*x]]) + (25*A*Cos[c + d*x]*Sin[c + d*x])/(12
*a*d*Sqrt[a - a*Sec[c + d*x]]) + (4*A*Cos[c + d*x]^2*Sin[c + d*x])/(3*a*d*Sqrt[a - a*Sec[c + d*x]])

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Rubi [A]  time = 0.701116, antiderivative size = 236, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {4020, 4022, 3920, 3774, 203, 3795} \[ \frac{85 A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a-a \sec (c+d x)}}\right )}{8 a^{3/2} d}-\frac{15 A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a-a \sec (c+d x)}}\right )}{\sqrt{2} a^{3/2} d}+\frac{35 A \sin (c+d x)}{8 a d \sqrt{a-a \sec (c+d x)}}+\frac{4 A \sin (c+d x) \cos ^2(c+d x)}{3 a d \sqrt{a-a \sec (c+d x)}}-\frac{A \sin (c+d x) \cos ^2(c+d x)}{d (a-a \sec (c+d x))^{3/2}}+\frac{25 A \sin (c+d x) \cos (c+d x)}{12 a d \sqrt{a-a \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^3*(A + A*Sec[c + d*x]))/(a - a*Sec[c + d*x])^(3/2),x]

[Out]

(85*A*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a - a*Sec[c + d*x]]])/(8*a^(3/2)*d) - (15*A*ArcTan[(Sqrt[a]*Tan[c + d
*x])/(Sqrt[2]*Sqrt[a - a*Sec[c + d*x]])])/(Sqrt[2]*a^(3/2)*d) - (A*Cos[c + d*x]^2*Sin[c + d*x])/(d*(a - a*Sec[
c + d*x])^(3/2)) + (35*A*Sin[c + d*x])/(8*a*d*Sqrt[a - a*Sec[c + d*x]]) + (25*A*Cos[c + d*x]*Sin[c + d*x])/(12
*a*d*Sqrt[a - a*Sec[c + d*x]]) + (4*A*Cos[c + d*x]^2*Sin[c + d*x])/(3*a*d*Sqrt[a - a*Sec[c + d*x]])

Rule 4020

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> -Simp[((A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(b*f*(2
*m + 1)), x] - Dist[1/(a^2*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[b*B*n - a*A*(2
*m + n + 1) + (A*b - a*B)*(m + n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*
b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0]

Rule 4022

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(f*n), x] - Dist[1
/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b*B*n - A*b*(m + n + 1)*Csc[e + f*x
], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[n, 0]

Rule 3920

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[c/a,
Int[Sqrt[a + b*Csc[e + f*x]], x], x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] /
; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0]

Rule 3774

Int[Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(-2*b)/d, Subst[Int[1/(a + x^2), x], x, (b*C
ot[c + d*x])/Sqrt[a + b*Csc[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 3795

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[-2/f, Subst[Int[1/(2
*a + x^2), x], x, (b*Cot[e + f*x])/Sqrt[a + b*Csc[e + f*x]]], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0
]

Rubi steps

\begin{align*} \int \frac{\cos ^3(c+d x) (A+A \sec (c+d x))}{(a-a \sec (c+d x))^{3/2}} \, dx &=-\frac{A \cos ^2(c+d x) \sin (c+d x)}{d (a-a \sec (c+d x))^{3/2}}+\frac{\int \frac{\cos ^3(c+d x) (8 a A+7 a A \sec (c+d x))}{\sqrt{a-a \sec (c+d x)}} \, dx}{2 a^2}\\ &=-\frac{A \cos ^2(c+d x) \sin (c+d x)}{d (a-a \sec (c+d x))^{3/2}}+\frac{4 A \cos ^2(c+d x) \sin (c+d x)}{3 a d \sqrt{a-a \sec (c+d x)}}-\frac{\int \frac{\cos ^2(c+d x) \left (-25 a^2 A-20 a^2 A \sec (c+d x)\right )}{\sqrt{a-a \sec (c+d x)}} \, dx}{6 a^3}\\ &=-\frac{A \cos ^2(c+d x) \sin (c+d x)}{d (a-a \sec (c+d x))^{3/2}}+\frac{25 A \cos (c+d x) \sin (c+d x)}{12 a d \sqrt{a-a \sec (c+d x)}}+\frac{4 A \cos ^2(c+d x) \sin (c+d x)}{3 a d \sqrt{a-a \sec (c+d x)}}+\frac{\int \frac{\cos (c+d x) \left (\frac{105 a^3 A}{2}+\frac{75}{2} a^3 A \sec (c+d x)\right )}{\sqrt{a-a \sec (c+d x)}} \, dx}{12 a^4}\\ &=-\frac{A \cos ^2(c+d x) \sin (c+d x)}{d (a-a \sec (c+d x))^{3/2}}+\frac{35 A \sin (c+d x)}{8 a d \sqrt{a-a \sec (c+d x)}}+\frac{25 A \cos (c+d x) \sin (c+d x)}{12 a d \sqrt{a-a \sec (c+d x)}}+\frac{4 A \cos ^2(c+d x) \sin (c+d x)}{3 a d \sqrt{a-a \sec (c+d x)}}-\frac{\int \frac{-\frac{255 a^4 A}{4}-\frac{105}{4} a^4 A \sec (c+d x)}{\sqrt{a-a \sec (c+d x)}} \, dx}{12 a^5}\\ &=-\frac{A \cos ^2(c+d x) \sin (c+d x)}{d (a-a \sec (c+d x))^{3/2}}+\frac{35 A \sin (c+d x)}{8 a d \sqrt{a-a \sec (c+d x)}}+\frac{25 A \cos (c+d x) \sin (c+d x)}{12 a d \sqrt{a-a \sec (c+d x)}}+\frac{4 A \cos ^2(c+d x) \sin (c+d x)}{3 a d \sqrt{a-a \sec (c+d x)}}+\frac{(85 A) \int \sqrt{a-a \sec (c+d x)} \, dx}{16 a^2}+\frac{(15 A) \int \frac{\sec (c+d x)}{\sqrt{a-a \sec (c+d x)}} \, dx}{2 a}\\ &=-\frac{A \cos ^2(c+d x) \sin (c+d x)}{d (a-a \sec (c+d x))^{3/2}}+\frac{35 A \sin (c+d x)}{8 a d \sqrt{a-a \sec (c+d x)}}+\frac{25 A \cos (c+d x) \sin (c+d x)}{12 a d \sqrt{a-a \sec (c+d x)}}+\frac{4 A \cos ^2(c+d x) \sin (c+d x)}{3 a d \sqrt{a-a \sec (c+d x)}}+\frac{(85 A) \operatorname{Subst}\left (\int \frac{1}{a+x^2} \, dx,x,\frac{a \tan (c+d x)}{\sqrt{a-a \sec (c+d x)}}\right )}{8 a d}-\frac{(15 A) \operatorname{Subst}\left (\int \frac{1}{2 a+x^2} \, dx,x,\frac{a \tan (c+d x)}{\sqrt{a-a \sec (c+d x)}}\right )}{a d}\\ &=\frac{85 A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a-a \sec (c+d x)}}\right )}{8 a^{3/2} d}-\frac{15 A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a-a \sec (c+d x)}}\right )}{\sqrt{2} a^{3/2} d}-\frac{A \cos ^2(c+d x) \sin (c+d x)}{d (a-a \sec (c+d x))^{3/2}}+\frac{35 A \sin (c+d x)}{8 a d \sqrt{a-a \sec (c+d x)}}+\frac{25 A \cos (c+d x) \sin (c+d x)}{12 a d \sqrt{a-a \sec (c+d x)}}+\frac{4 A \cos ^2(c+d x) \sin (c+d x)}{3 a d \sqrt{a-a \sec (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 6.68658, size = 452, normalized size = 1.92 \[ A \left (\frac{\sin ^3\left (\frac{c}{2}+\frac{d x}{2}\right ) \sec ^2(c+d x) \left (\frac{65 \sin \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right )}{12 d}+\frac{25 \sin \left (\frac{3 c}{2}\right ) \sin \left (\frac{3 d x}{2}\right )}{3 d}+\frac{5 \sin \left (\frac{5 c}{2}\right ) \sin \left (\frac{5 d x}{2}\right )}{4 d}+\frac{\sin \left (\frac{7 c}{2}\right ) \sin \left (\frac{7 d x}{2}\right )}{6 d}-\frac{65 \cos \left (\frac{c}{2}\right ) \cos \left (\frac{d x}{2}\right )}{12 d}-\frac{25 \cos \left (\frac{3 c}{2}\right ) \cos \left (\frac{3 d x}{2}\right )}{3 d}-\frac{5 \cos \left (\frac{5 c}{2}\right ) \cos \left (\frac{5 d x}{2}\right )}{4 d}-\frac{\cos \left (\frac{7 c}{2}\right ) \cos \left (\frac{7 d x}{2}\right )}{6 d}-\frac{2 \cot \left (\frac{c}{2}\right ) \csc \left (\frac{c}{2}+\frac{d x}{2}\right )}{d}+\frac{2 \csc \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right ) \csc ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{d}\right )}{(a-a \sec (c+d x))^{3/2}}-\frac{5 e^{-\frac{1}{2} i (c+d x)} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \sin ^3\left (\frac{c}{2}+\frac{d x}{2}\right ) \sec ^{\frac{3}{2}}(c+d x) \left (17 \sinh ^{-1}\left (e^{i (c+d x)}\right )-24 \sqrt{2} \tanh ^{-1}\left (\frac{1+e^{i (c+d x)}}{\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}}\right )+17 \tanh ^{-1}\left (\sqrt{1+e^{2 i (c+d x)}}\right )\right )}{4 \sqrt{2} d (a-a \sec (c+d x))^{3/2}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^3*(A + A*Sec[c + d*x]))/(a - a*Sec[c + d*x])^(3/2),x]

[Out]

A*((-5*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*(17*ArcSinh[E^(I*(c + d*x
))] - 24*Sqrt[2]*ArcTanh[(1 + E^(I*(c + d*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))])] + 17*ArcTanh[Sqrt[1 +
E^((2*I)*(c + d*x))]])*Sec[c + d*x]^(3/2)*Sin[c/2 + (d*x)/2]^3)/(4*Sqrt[2]*d*E^((I/2)*(c + d*x))*(a - a*Sec[c
+ d*x])^(3/2)) + (Sec[c + d*x]^2*((-65*Cos[c/2]*Cos[(d*x)/2])/(12*d) - (25*Cos[(3*c)/2]*Cos[(3*d*x)/2])/(3*d)
- (5*Cos[(5*c)/2]*Cos[(5*d*x)/2])/(4*d) - (Cos[(7*c)/2]*Cos[(7*d*x)/2])/(6*d) - (2*Cot[c/2]*Csc[c/2 + (d*x)/2]
)/d + (2*Csc[c/2]*Csc[c/2 + (d*x)/2]^2*Sin[(d*x)/2])/d + (65*Sin[c/2]*Sin[(d*x)/2])/(12*d) + (25*Sin[(3*c)/2]*
Sin[(3*d*x)/2])/(3*d) + (5*Sin[(5*c)/2]*Sin[(5*d*x)/2])/(4*d) + (Sin[(7*c)/2]*Sin[(7*d*x)/2])/(6*d))*Sin[c/2 +
(d*x)/2]^3)/(a - a*Sec[c + d*x])^(3/2))

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Maple [B]  time = 0.326, size = 1104, normalized size = 4.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(A+A*sec(d*x+c))/(a-a*sec(d*x+c))^(3/2),x)

[Out]

1/168*A/d*2^(1/2)*(-1+cos(d*x+c))^5*(1130*cos(d*x+c)^3*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-3570*arcta
n(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+720*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(7/2)*cos(d*x+c)*2^(1/2
)+1008*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*cos(d*x+c)^3*2^(1/2)-1680*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*cos
(d*x+c)^3*2^(1/2)+5040*arctan(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)^3*2^(1/2)-720*(-2*cos(d*x+c)/
(cos(d*x+c)+1))^(7/2)*cos(d*x+c)^3*2^(1/2)-504*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*2^(1/2)-735*2^(1/2)*(-2*co
s(d*x+c)/(cos(d*x+c)+1))^(1/2)-2520*2^(1/2)*arctan(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))-1008*(-2*cos(d*x+c)
/(cos(d*x+c)+1))^(5/2)*cos(d*x+c)*2^(1/2)+1225*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)^5*2^(1/2)-2103*
(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)^4*2^(1/2)-168*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(9/2)+357
0*cos(d*x+c)^4*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+1680*cos(d*x+c)*2^(1/2)*(-2*cos(d*x+c)
/(cos(d*x+c)+1))^(3/2)+952*cos(d*x+c)^2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-875*cos(d*x+c)*2^(1/2)*(-
2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-5040*cos(d*x+c)*2^(1/2)*arctan(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+840*
2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)-7140*cos(d*x+c)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))
^(1/2))+360*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(7/2)*2^(1/2)+7140*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1)
)^(1/2))*cos(d*x+c)^3-672*2^(1/2)*cos(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(9/2)+504*2^(1/2)*cos(d*x+c)^4*(-2
*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)-840*2^(1/2)*cos(d*x+c)^4*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)+2520*2^(1/2)*c
os(d*x+c)^4*arctan(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+56*2^(1/2)*cos(d*x+c)^7*(-2*cos(d*x+c)/(cos(d*x+c)+
1))^(1/2)+350*2^(1/2)*cos(d*x+c)^6*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-168*2^(1/2)*cos(d*x+c)^4*(-2*cos(d*x+c
)/(cos(d*x+c)+1))^(9/2)-672*2^(1/2)*cos(d*x+c)^3*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(9/2)-360*2^(1/2)*cos(d*x+c)^4
*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(7/2)-1008*2^(1/2)*cos(d*x+c)^2*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(9/2))/(-2*cos(
d*x+c)/(cos(d*x+c)+1))^(3/2)/(a*(-1+cos(d*x+c))/cos(d*x+c))^(3/2)/sin(d*x+c)^9

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (A \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{3}}{{\left (-a \sec \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(A+A*sec(d*x+c))/(a-a*sec(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate((A*sec(d*x + c) + A)*cos(d*x + c)^3/(-a*sec(d*x + c) + a)^(3/2), x)

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Fricas [A]  time = 0.572233, size = 1490, normalized size = 6.31 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(A+A*sec(d*x+c))/(a-a*sec(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

[-1/48*(180*sqrt(2)*(A*cos(d*x + c) - A)*sqrt(-a)*log((2*sqrt(2)*(cos(d*x + c)^2 + cos(d*x + c))*sqrt(-a)*sqrt
((a*cos(d*x + c) - a)/cos(d*x + c)) + (3*a*cos(d*x + c) + a)*sin(d*x + c))/((cos(d*x + c) - 1)*sin(d*x + c)))*
sin(d*x + c) + 255*(A*cos(d*x + c) - A)*sqrt(-a)*log((2*(cos(d*x + c)^2 + cos(d*x + c))*sqrt(-a)*sqrt((a*cos(d
*x + c) - a)/cos(d*x + c)) - (2*a*cos(d*x + c) + a)*sin(d*x + c))/sin(d*x + c))*sin(d*x + c) + 2*(8*A*cos(d*x
+ c)^5 + 26*A*cos(d*x + c)^4 + 73*A*cos(d*x + c)^3 - 50*A*cos(d*x + c)^2 - 105*A*cos(d*x + c))*sqrt((a*cos(d*x
+ c) - a)/cos(d*x + c)))/((a^2*d*cos(d*x + c) - a^2*d)*sin(d*x + c)), 1/24*(180*sqrt(2)*(A*cos(d*x + c) - A)*
sqrt(a)*arctan(sqrt(2)*sqrt((a*cos(d*x + c) - a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c)))*sin(d*x +
c) - 255*(A*cos(d*x + c) - A)*sqrt(a)*arctan(sqrt((a*cos(d*x + c) - a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin
(d*x + c)))*sin(d*x + c) - (8*A*cos(d*x + c)^5 + 26*A*cos(d*x + c)^4 + 73*A*cos(d*x + c)^3 - 50*A*cos(d*x + c)
^2 - 105*A*cos(d*x + c))*sqrt((a*cos(d*x + c) - a)/cos(d*x + c)))/((a^2*d*cos(d*x + c) - a^2*d)*sin(d*x + c))]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**3*(A+A*sec(d*x+c))/(a-a*sec(d*x+c))**(3/2),x)

[Out]

Timed out

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Giac [A]  time = 2.23083, size = 425, normalized size = 1.8 \begin{align*} -\frac{A{\left (\frac{180 \, \sqrt{2} \arctan \left (\frac{\sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a}}{\sqrt{a}}\right )}{a^{\frac{3}{2}} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right ) \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )} - \frac{255 \, \arctan \left (\frac{\sqrt{2} \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a}}{2 \, \sqrt{a}}\right )}{a^{\frac{3}{2}} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right ) \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )} - \frac{\sqrt{2}{\left (63 \,{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )}^{\frac{5}{2}} + 272 \,{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )}^{\frac{3}{2}} a + 324 \, \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a} a^{2}\right )}}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a\right )}^{3} a \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right ) \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )} - \frac{12 \, \sqrt{2} \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a}}{a^{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right ) \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right ) \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}\right )}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(A+A*sec(d*x+c))/(a-a*sec(d*x+c))^(3/2),x, algorithm="giac")

[Out]

-1/24*A*(180*sqrt(2)*arctan(sqrt(a*tan(1/2*d*x + 1/2*c)^2 - a)/sqrt(a))/(a^(3/2)*sgn(tan(1/2*d*x + 1/2*c)^2 -
1)*sgn(tan(1/2*d*x + 1/2*c))) - 255*arctan(1/2*sqrt(2)*sqrt(a*tan(1/2*d*x + 1/2*c)^2 - a)/sqrt(a))/(a^(3/2)*sg
n(tan(1/2*d*x + 1/2*c)^2 - 1)*sgn(tan(1/2*d*x + 1/2*c))) - sqrt(2)*(63*(a*tan(1/2*d*x + 1/2*c)^2 - a)^(5/2) +
272*(a*tan(1/2*d*x + 1/2*c)^2 - a)^(3/2)*a + 324*sqrt(a*tan(1/2*d*x + 1/2*c)^2 - a)*a^2)/((a*tan(1/2*d*x + 1/2
*c)^2 + a)^3*a*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)*sgn(tan(1/2*d*x + 1/2*c))) - 12*sqrt(2)*sqrt(a*tan(1/2*d*x + 1/
2*c)^2 - a)/(a^2*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)*sgn(tan(1/2*d*x + 1/2*c))*tan(1/2*d*x + 1/2*c)^2))/d